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hughwill
Joined: 05 Apr 2010
Posts: 424
Location: Birmingham UK
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 Posted: Wed May 18, 2016 8:47 am Post subject: May 18 VH |
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After basics: | Code: |
+------------+------------+--------------+
| 4 12 1269 | 8 7 569 | 3 59 169 |
| 58 7 169 | 3 569 2 | 158 4 169 |
| 58 3 69 | 56 4 1 | 78 579 2 |
+------------+------------+--------------+
| 27 12 127 | 56 56 4 | 9 8 3 |
| 9 4 8 | 2 3 7 | 6 1 5 |
| 3 6 5 | 19 18 89 | 47 2 47 |
+------------+------------+--------------+
| 1 8 24 | 7 569 569 | 245 3 49 |
| 27 9 3 | 4 18 58 | 1257 6 17 |
| 6 5 47 | 19 2 3 | 147 79 8 |
+------------+------------+--------------+
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Play this puzzle online at the Daily Sudoku site
This time a VH which is (by the rules) a true VH: Either | Quote: | | 79-4XY wing pivot r9c8 or 49-7 XY-Wing pivot r7c9 |
will solve this in one. (Select white area to view solution)
Hugh |
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Marty R.
Joined: 12 Feb 2006
Posts: 5770
Location: Rochester, NY, USA
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| Posted: Wed May 18, 2016 10:57 pm Post subject: |
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| yep, beats a basics-only solution. |
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dongrave
Joined: 06 Mar 2014
Posts: 568
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| Posted: Thu May 19, 2016 12:59 am Post subject: |
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| I had a couple more basic moves in Hugh's grid. The locked 5's in column 8 of box 3 removes it from r2c7 and then that gives you a hidden 25 pair in box 9. Then I found the 147 XYZ Wing with pivot at r9c7 (pincers at r9c3 and r8c9) so r9c8<>7 which solves it. |
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Marty R.
Joined: 12 Feb 2006
Posts: 5770
Location: Rochester, NY, USA
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| Posted: Thu May 19, 2016 1:37 pm Post subject: |
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| dongrave wrote: | | I had a couple more basic moves in Hugh's grid. The locked 5's in column 8 of box 3 removes it from r2c7 and then that gives you a hidden 25 pair in box 9. Then I found the 147 XYZ Wing with pivot at r9c7 (pincers at r9c3 and r8c9) so r9c8<>7 which solves it. |
I don't understand. Why do locked 5s matter? The hidden pair on 25 (or as I prefer, a 1479 quad ) is already there. After finishing basics, there's no elimination from the XYZ-Wing that I can see. |
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dongrave
Joined: 06 Mar 2014
Posts: 568
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| Posted: Thu May 19, 2016 3:23 pm Post subject: |
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| Marty R. wrote: | | dongrave wrote: | | I had a couple more basic moves in Hugh's grid. The locked 5's in column 8 of box 3 removes it from r2c7 and then that gives you a hidden 25 pair in box 9. Then I found the 147 XYZ Wing with pivot at r9c7 (pincers at r9c3 and r8c9) so r9c8<>7 which solves it. |
I don't understand. Why do locked 5s matter? The hidden pair on 25 (or as I prefer, a 1479 quad ) is already there. After finishing basics, there's no elimination from the XYZ-Wing that I can see. |
Hi Marty, It's not that the locked 5's matter; it's just another basic move that I saw. And after that, I saw that there was a hidden 25 pair so I pointed that out too. I think your point is that my phrasing of 'that gives you' is wrong though because the hidden pair is there already - but I didn't see it until I removed the 5 from r2c7. I posted what the XYZ wing does because I think it's easy to miss (Maybe you're not looking at the right XYZ Wing?) The 147 XYZ with pivot at r9c7 and pincers at r9c3 and r8c9 all have a 7 in them and they are all buddies with r9c8 so it can't be 7. Am I missing something? |
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George Woods
Joined: 28 Mar 2006
Posts: 304
Location: Dorset UK
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| Posted: Thu May 19, 2016 3:50 pm Post subject: a different xy |
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| nobody mentioned the 179 XY pivot r9c8 |
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Marty R.
Joined: 12 Feb 2006
Posts: 5770
Location: Rochester, NY, USA
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| Posted: Thu May 19, 2016 3:52 pm Post subject: |
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| Quote: | | (Maybe you're not looking at the right XYZ Wing?) |
That is correct.   |
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dongrave
Joined: 06 Mar 2014
Posts: 568
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| Posted: Thu May 19, 2016 3:57 pm Post subject: |
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| Marty R. wrote: | | Quote: | | (Maybe you're not looking at the right XYZ Wing?) |
That is correct. |
No big deal Marty! You're allowed to make mistakes! (It was't nearly as obvious as what I missed!) And I completely missed George's XY wing - and I looked for other wings for a while! |
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